How do you integrate #int 1/(xsqrtx)dx#?

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Answer 1 · 2017-02-16T12:54:33

The answer is #=-2/sqrtx+C#

We use

#intx^ndx=x^(n+1)/(n+1)+C (n!=-1)#

#sqrtx=x^(1/2)#

Therefore,

#intdx/(xsqrtx)#

#=intdx/(x^(3/2))#

#=intx^(-3/2)dx#

#=x^(-3/2+1)/(-3/2+1)+C#

#=-2x^(-1/2)+C#

#=-2/sqrtx+C#

Answer 2 · 2017-02-16T12:58:30

#-2/sqrt x +C#

Use power rule to integrate as shown below:

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